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Proof of non-existence of perfect cuboids

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Let $a$ , $b$ , $c$ , $d$ , $e$ , $f$ and $g$ be positive integers. We consider that $a$ , $b$ and $c$ have no common prime factors and $a$ , $b$ and $c$ are the edge lengths, $d$ , $e$ and $f$ are the diagonal lengths and $g$ is the space diagonal length. If a perfect cuboid exists, the following equations hold.
$a^{2} + b^{2} = d^{2}$
$b^{2} + c^{2} = e^{2}$
$c^{2} + a^{2} = f^{2}$
$a^{2} + b^{2} + c^{2} = g^{2}$
$a$ is odd and $b$ and $c$ are even because $a$ , $b$ and $c$ are not all even and if two edges are odd, then the diagonal is not a square number. And $d$ and $f$ are odd and $e$ is even. Let $i$ , $k_{i}$ , $m_{i}$ and $n_{i}$ be positive inetgers and $1 ≦ i ≦ 6$ hods. We consider that $m_{i}$ and $n_{i}$ are relatively prime and the odd/even number of $m_{i}$ and $n_{i}$ are different. By the form of Pythagorean triple, the following equations hold.
$a = (m_{1}^{2} - n_{1}^{2}) k_{1} = (m_{2}^{2} - n_{2}^{2}) k_{2} = (m_{5}^{2} - n_{5}^{2}) k_{5}$
$b = 2 k_{2} m_{2} n_{2} = (m_{3}^{2} - n_{3}^{2}) k_{3} = 2 k_{6} m_{6} n_{6}$
$c = 2 k_{3} m_{3} n_{3} = 2 k_{1} m_{1} n_{1} = 2 k_{4} m_{4} n_{4}$
$d = (m_{2}^{2} + n_{2}^{2}) k_{2} = (m_{4}^{2} - n_{4}^{2}) k_{4}$
$e = (m_{3}^{2} + n_{3}^{2}) k_{3} = 2 k_{5} m_{5} n_{5}$
$f = (m_{1}^{2} + n_{1}^{2}) k_{1} = (m_{6}^{2} - n_{6}^{2}) k_{6}$
$g = (m_{4}^{2} + n_{4}^{2}) k_{4} = (m_{5}^{2} + n_{5}^{2}) k_{5} = (m_{6}^{2} + n_{6}^{2}) k_{6}$
$k_{3}$ is even and the others $k_{i}$ are all odd. We consider that $k_{4}$ , $k_{5}$ and $k_{6}$ are each prime to each other since the greatest common divisor of $a$ , $b$ and $c$ is one. Let $r$ be an odd positive integer. Since $g$ has factors $k_{4}$ , $k_{5}$ and $k_{6}$ ,
$g^{2} = ((m_{5}^{2} - n_{5}^{2}) k_{5})^{2} + (2 k_{6} m_{6} n_{6})^{2} + (2 k_{4} m_{4} n_{4})^{2} = (r k_{4} k_{5} k_{6})^{2} . . . (1)$
holds. Let $s$ and $u$ be odd positive integers and $t$ be an even positive integer. By this equation, the following equations hold.
$((m_{5}^{2} - n_{5}^{2}) k_{5})^{2} + (2 k_{6} m_{6} n_{6})^{2} = s k_{4}^{2}$
$(2 k_{6} m_{6} n_{6})^{2} + (2 k_{4} m_{4} n_{4})^{2} = t k_{5}^{2}$
$(2 k_{4} m_{4} n_{4})^{2} + ((m_{5}^{2} - n_{5}^{2}) k_{5})^{2} = u k_{6}^{2}$
By these equations,
$s k_{4}^{2} + t k_{5}^{2} + u k_{6}^{2} = 2 (r k_{4} k_{5} k_{6})^{2}$
holds. Let $v$ and $w$ be odd positive integers and $x$ be an even positive integer. By this equation, the following equations hold.
$s k_{4}^{2} + t k_{5}^{2} = v k_{6}^{2}$
$t k_{5}^{2} + u k_{6}^{2} = w k_{4}^{2}$
$u k_{6}^{2} + s k_{4}^{2} = x k_{5}^{2}$
By these equations,
$w k_{4}^{2} + x k_{5}^{2} + v k_{6}^{2} = (2 r k_{4} k_{5} k_{6})^{2} . . . (2)$
holds. If $v$ has a common prime factor of $k_{4}$ , then $t$ have the common prime factor and $u$ and $x$ have it also. In this case, $d$ , $e$ and $f$ have the common prime factor and this contradicts that $a$ , $b$ and $c$ have no common prime factors. Therefore, $v$ and $k_{4}$ are coprime. In the same way, $v$ and $k_{5}$ , $w$ and $k_{5}$ , $w$ and $k_{6}$ , $x$ and $k_{6}$ and $x$ and $k_{4}$ are relatively prime. Comparing the equation (2) with the equation (1), we find that the following equations hold since there exists a perfect cuboid whose three sides are twice as long as the edge lengths of the equation (1).
$v = (4 m_{6} n_{6})^{2}$
$w = (4 m_{4} n_{4})^{2}$
$x = (2 (m_{5}^{2} - n_{5}^{2}))^{2}$
It becomes a contradiction since $v$ and $w$ are odd. From the above, there are no perfect cuboids. (Q.E.D.)